Correct Answer - Option 2 : 3.6 × 10
-22 kg. ms
-1
Concept:
Variation of Mass with Velocity
According to classical physics, the inertial mass of a body is independent of the velocity of light. It is regarding as a constant. However special theory of relativity leads us to the concept of variation of mass with velocity. It follows from special theory of relativity that the mass m of a body moving with relativistic velocity v relative to an observer is larger than its m0 when it is at rest.
According to Einstein, the mass of the body in motion is different from the mass of the body at rest.
\(m = \frac{{{m_o}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)
This is the relative formula for variation of mass with velocity where m0 is the rest mass and m is the relativistic mass of the body.
Realistic Kinetic Energy:
The kinetic energy of a particle of rest mass m0 where m is the mass of the particle when it is moving with a velocity c is:
\({E_k} = \left( {m - {m_o}} \right){c^2} = {m_o}{c^2}\left[ {\frac{1}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} - 1} \right]\)
When v < c
\(\begin{array}{l} {\left( {1 - \frac{{{v^2}}}{{{c^2}}}} \right)^{ - \frac{1}{2}}} \approx 1 + \frac{1}{2}\frac{{{v^2}}}{{{c^2}}}\\ \therefore {E_k} = \left( {m - {m_o}} \right){c^2} = {m_o}{c^2}\left( {\frac{1}{2}\frac{{{v^2}}}{{{c^2}}}} \right) = \frac{1}{2}{m_o}{v^2} \end{array}\)
Total energy:
\(E = {E_k} + {m_0}{c^2}\)
Where Ek is the kinetic energy and m0c2 the rest mass energy.
\(E = m{c^2} = \frac{{{m_o}{c^2}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}\)
Megaelectronvolt (MeV)
1 MeV = 106 eV
1 eV = 1.6×10−19 joules
Calculation:
The momentum of the given electron = Apparent mass × given velocity
\(\begin{array}{l} \frac{{{m_0}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }} \times v\\ = \frac{{9.11 \times {{10}^{ - 31}}}}{{\sqrt {1 - {{\left( {0.8} \right)}^2}} }} \times 0.8 \times 3 \times {10^8} \end{array}\)
= 3.6 × 10
-22 kg m/s
